You intend to estimate a population mean mu with the following sample.

36.1
89.4
69.1
61.9
104.2
41.4
67.1
54.6
90.2
68.9

You believe the population is normally distributed. Find the 95% confidence interval. Enter your answer as an open-interval (i.e., parentheses) accurate to two decimal places (because the sample data are reported accurate to one decimal place).

95% C.I. =

Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.

1 answer

To calculate a 95% confidence interval for the population mean \(\mu\) based on the given sample data, we can follow these steps:

  1. Calculate the sample mean (\(\bar{x}\)).
  2. Calculate the sample standard deviation (\(s\)).
  3. Determine the sample size (\(n\)).
  4. Determine the critical value (\(t^*\)) for a 95% confidence interval using the t-distribution.
  5. Calculate the margin of error.
  6. Construct the confidence interval.

Let's go through each step.

Step 1: Calculate the sample mean

\[ \bar{x} = \frac{\sum x_i}{n} \]

Given the sample data:

  • \(x_1 = 36.1\)
  • \(x_2 = 89.4\)
  • \(x_3 = 69.1\)
  • \(x_4 = 61.9\)
  • \(x_5 = 104.2\)
  • \(x_6 = 41.4\)
  • \(x_7 = 67.1\)
  • \(x_8 = 54.6\)
  • \(x_9 = 90.2\)
  • \(x_{10} = 68.9\)

Calculating the total sum:

\[ \sum x_i = 36.1 + 89.4 + 69.1 + 61.9 + 104.2 + 41.4 + 67.1 + 54.6 + 90.2 + 68.9 = 683.8 \]

The sample size \(n\) is 10.

Thus, the sample mean is:

\[ \bar{x} = \frac{683.8}{10} = 68.38 \]

Step 2: Calculate the sample standard deviation

The formula for the sample standard deviation is:

\[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}} \]

Calculating \( (x_i - \bar{x})^2 \):

  1. \( (36.1 - 68.38)^2 = 1032.0644 \)
  2. \( (89.4 - 68.38)^2 = 444.0644 \)
  3. \( (69.1 - 68.38)^2 = 0.5184 \)
  4. \( (61.9 - 68.38)^2 = 41.5844 \)
  5. \( (104.2 - 68.38)^2 = 1294.9444 \)
  6. \( (41.4 - 68.38)^2 = 724.0644 \)
  7. \( (67.1 - 68.38)^2 = 1.6324 \)
  8. \( (54.6 - 68.38)^2 = 188.0644 \)
  9. \( (90.2 - 68.38)^2 = 473.1844 \)
  10. \( (68.9 - 68.38)^2 = 0.2724 \)

Now summing those squared differences:

\[ \sum (x_i - \bar{x})^2 = 1032.0644 + 444.0644 + 0.5184 + 41.5844 + 1294.9444 + 724.0644 + 1.6324 + 188.0644 + 473.1844 + 0.2724 = 3438.25 \]

Then we can find \(s\):

\[ s = \sqrt{\frac{3438.25}{10 - 1}} = \sqrt{\frac{3438.25}{9}} = \sqrt{382.0278} \approx 19.55 \]

Step 3: Determine the sample size

The sample size \(n = 10\).

Step 4: Determine the critical value \(t^*\)

Using the t-distribution for 95% confidence level and \(n - 1 = 9\) degrees of freedom, we can find \(t^*\) from a t-table or calculator.

For \(df = 9\):

\(t^* \approx 2.262\) (rounded to three decimal places).

Step 5: Calculate the margin of error

The margin of error (ME) is given by:

\[ ME = t^* \times \frac{s}{\sqrt{n}} \]

Calculating the margin of error:

\[ ME = 2.262 \times \frac{19.55}{\sqrt{10}} \approx 2.262 \times \frac{19.55}{3.162} \approx 2.262 \times 6.182 \approx 13.96 \]

Step 6: Construct the confidence interval

To find the confidence interval, we calculate:

\[ \bar{x} - ME \quad \text{and} \quad \bar{x} + ME \]

\[ C.I. = (68.38 - 13.96, 68.38 + 13.96) = (54.42, 82.34) \]

Thus, the final answer for the 95% confidence interval is:

\[ \text{95% C.I.} = (54.42, 82.34) \]

So, the answer is:

\[ \text{95% C.I.} = (54.42, 82.34) \]